Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 3)
3.
A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
Answer: Option
Explanation:
| A's 2 day's work = |  |
1 | x 2 |  |
= | 1 | . |
| 20 | 10 |
| (A + B + C)'s 1 day's work = | |
1 | + | 1 | + | 1 | |
= | 6 | = | 1 | . |
| 20 | 30 | 60 | 60 | 10 |
| Work done in 3 days = | |
1 | + | 1 | |
= | 1 | . |
| 10 | 10 | 5 |
| Now, | 1 | work is done in 3 days. |
| 5 |
Whole work will be done in (3 x 5) = 15 days.
Discussion:
357 comments Page 14 of 36.
ROSE said:
1 decade ago
Work done in 3days=1/10(A'S 2DAYS WORK)+1/10(1DAY WORK FROM ABC)
=2/10=1/5
NOW CROSS MULTIPLY FOR 1/5 3DAYS ARE REQUIRED FOR 5/5 DAYS HOW MANY DAYS ARE REQUIRED
NOW 5/5=1
1*3/1/5=
3*5/1=15
=2/10=1/5
NOW CROSS MULTIPLY FOR 1/5 3DAYS ARE REQUIRED FOR 5/5 DAYS HOW MANY DAYS ARE REQUIRED
NOW 5/5=1
1*3/1/5=
3*5/1=15
PARSHU said:
9 years ago
3 values there 3persons working to gather assisted A's B + C we calculate A + B + C/2 we get,
A+ B + C = 1/20 + 1/30 + 1/60.
LCM denomineor = 1/10 = 10 * 3 total 30/2 = 15-> 3persons work.
A+ B + C = 1/20 + 1/30 + 1/60.
LCM denomineor = 1/10 = 10 * 3 total 30/2 = 15-> 3persons work.
Abishek Aryal said:
7 months ago
In 1 day A can do 1/20.
In 3 days A can do 3/20.
In 1 day B can do 1/30.
In 1 day c can da 1/60
In 3 days ( 3 day of A+ 1 day of B + 1 day of C) = 1/5.
Now,
1/5 in 3 days.
1 in 15 days.
In 3 days A can do 3/20.
In 1 day B can do 1/30.
In 1 day c can da 1/60
In 3 days ( 3 day of A+ 1 day of B + 1 day of C) = 1/5.
Now,
1/5 in 3 days.
1 in 15 days.
(51)
Anurag said:
1 decade ago
Let N be the total no.of days A has to work. So, a/q B and C works only for N/3 days. Total work = 1.
Therefore,
N*(1/20)+ (N/3)*((1/30)+(1/60)) = 1.
Solving this , you will get N = 15.
Therefore,
N*(1/20)+ (N/3)*((1/30)+(1/60)) = 1.
Solving this , you will get N = 15.
A.Venkatesh Reddy said:
6 years ago
A=20 A Effi=3.
B=30 LCM=60 B Effi=2.
So,Work=60 => 3+3+6=12(3 days work)=> 12*5=60=>15days.
C=60 C Effi=1.
B=30 LCM=60 B Effi=2.
So,Work=60 => 3+3+6=12(3 days work)=> 12*5=60=>15days.
C=60 C Effi=1.
Narayana said:
1 decade ago
A work in 2 Days = 1/20*2 = 1/10.
After 3rd day add in B, C. So A, B, C work in 1 day.
= 1/20+1/30+1/60 = 1/10.
2days + 1 day = 1/10+1/10 => 1/5.
Work completed = 1/5*3 = 15 days.
After 3rd day add in B, C. So A, B, C work in 1 day.
= 1/20+1/30+1/60 = 1/10.
2days + 1 day = 1/10+1/10 => 1/5.
Work completed = 1/5*3 = 15 days.
Rah said:
9 years ago
Suppose B and C are helping a, = b 1/30 + c 1/60 + a 1/20.
But B and C are helping only on third day so( b+c / 3)+a = (1/30+1/60 =3/60 = 1/20/3=1/60)+ 1/20= 1/20+1/60=4/60=1/15 = 15.
But B and C are helping only on third day so( b+c / 3)+a = (1/30+1/60 =3/60 = 1/20/3=1/60)+ 1/20= 1/20+1/60=4/60=1/15 = 15.
Veeresh said:
7 years ago
@All.
A is 1st-day work 1/20.
A is 2nd-days work 1/2.
02/20.
a+b+c together work is = 1/20 + 1/30 + 1/60.
Add 2/20+1/20+1/30+1/60 = 1/5.
For 3 days = 1/5 * 1/3 = 1/15 ie 15 days.
A is 1st-day work 1/20.
A is 2nd-days work 1/2.
02/20.
a+b+c together work is = 1/20 + 1/30 + 1/60.
Add 2/20+1/20+1/30+1/60 = 1/5.
For 3 days = 1/5 * 1/3 = 1/15 ie 15 days.
16cs1045 said:
6 years ago
10 %for A in first 2 day 10%(5+3.33+1.66)5 for A in third day and 5 for B and C in third similarly for other 12 day work will complete. Therefore the total time required is 15 day.
Shakti said:
1 decade ago
Let total work = 60.
A's effi.=3, same as B and C have 2 and 1.
Then work
1st 2nd 3rd
3 3 6
Total work in 3 days 12 and hence they take 15 days to complete the work(60).
A's effi.=3, same as B and C have 2 and 1.
Then work
1st 2nd 3rd
3 3 6
Total work in 3 days 12 and hence they take 15 days to complete the work(60).
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